Question: $\lim_{x\to 3}\dfrac{x-3}{2-\sqrt{x+1}}=$
Answer: Substituting $x=3$ into $\dfrac{x-3}{2-\sqrt{x+1}}$ results in the indeterminate form $\dfrac{0}{0}$. This doesn't necessarily mean the limit doesn't exist, but it does mean we have to work a little before we find it. Since we have a rational expression with a square root on our hands, let's try to re-write it using the method of rationalization. $\begin{aligned} &\phantom{=}\dfrac{x-3}{2-\sqrt{x+1}} \\\\ &=\dfrac{x-3}{2-\sqrt{x+1}}\cdot\dfrac{2+\sqrt{x+1}}{2+\sqrt{x+1}} \gray{\text{Rationalize the denominator}} \\\\ &=\dfrac{(x-3)(2+\sqrt{x+1})}{2^2-(x+1)} \\\\ &=\dfrac{\cancel{(x-3)}(2+\sqrt{x+1})}{-1\cancel{(x-3)}} \gray{\text{Cancel out common factors}} \\\\ &=-2-\sqrt{x+1} \text{, for }x\neq 3 \end{aligned}$ This means that the two expressions have the same value for all $x$ -values (in their domains) except for $3$. We can now use the following theorem: If $f(x)=g(x)$ for all $x$ -values in a given interval except for $x=c$, then $\lim_{x\to c}f(x)=\lim_{x\to c}g(x)$. In our case, $\dfrac{x-3}{2-\sqrt{x+1}}=-2-\sqrt{x+1}$ for all $x$ -values in the interval $(2.5,3.5)$ except for $x=3$. Therefore, $\lim_{x\to 3}\dfrac{x-3}{2-\sqrt{x+1}}=\lim_{x\to 3}-2-\sqrt{x+1}=-4$. (The last limit was found using direct substitution.) [I want to see how this looks graphically!] In conclusion, $\lim_{x\to 3}\dfrac{x-3}{2-\sqrt{x+1}}=-4$.